We note first that division by zero is not allowed and that for this reason \(8x-7\) may not be equal to zero and that therefore \(x={{7}\over{8}}\) is not a solution.

We now distinguish two cases, namely \(8x-7>0\) and \(8x-7<0\).
In both cases we multiply the inequality on both sides by \(8x-7\) because we then get a linear inequality, for which we know there is a solution method.

Suppose \(8x-7>0\), i.e. \(x> {{7}\over{8}}\). Then we get \(4<-(8x-7)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(8x<3\).
Then, dvision by the coefficient of \(x\)gives \(x < {{3}\over{8}}\).
So we have the following system of inequalities: \(x> {{7}\over{8}}\,\wedge\; x < {{3}\over{8}}\)
and this simplifies to \(\text{an empty solution set}\).

Suppose \(8x-7<0\), i.e. \(x< {{7}\over{8}}\). Then we get \(4>-(8x-7)\).
When we move everything with \(x\) to the left and all constant terms to the right, we get \(8x>3\).
Then, division by the coefficient of \(x\) gives \(x > {{3}\over{8}}\).
So we have the following system of inequalities: \(x< {{7}\over{8}}\,\wedge\; x > {{3}\over{8}}\)
and this simplifies to \({{3}\over{8}}\lt x\lt {{7}\over{8}}\).

The solution of the original inequality is \({{3}\over{8}}\lt x\lt {{7}\over{8}}\).